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102. Binary Tree Level Order Traversal

时间:2016-06-10 06:12:49      阅读:167      评论:0      收藏:0      [点我收藏+]

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BSF用queue,DSF用stack

 

 1 public List<List<Integer>> levelOrder(TreeNode root) {
 2         List<List<Integer>> res = new ArrayList<List<Integer>>();
 3         if(root == null) {
 4             return res;
 5         }
 6         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
 7         queue.offer(root);
 8         int lastLevelCnt = 1;
 9         int curLevelCnt = 0;
10         List<Integer> curLevel = new ArrayList<Integer>();
11         while(!queue.isEmpty()) {
12             TreeNode curNode = queue.poll();
13             curLevel.add(curNode.val);
14             lastLevelCnt--;
15             if(curNode.left != null) {
16                 queue.add(curNode.left);
17                 curLevelCnt++;
18             }
19             if(curNode.right != null) {
20                 queue.add(curNode.right);
21                 curLevelCnt++;
22                 
23             }
24             if(lastLevelCnt == 0) {
25                 res.add(curLevel);
26                 lastLevelCnt = curLevelCnt;
27                 curLevelCnt = 0;
28                 curLevel = new ArrayList<Integer>();
29             }
30         }
31         return res;
32     }

 

102. Binary Tree Level Order Traversal

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原文地址:http://www.cnblogs.com/warmland/p/5573077.html

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